We have to find the first and second derivative of f(x)=x^(2/3)(6-x)^(1/3)
I have the first derivative as being
(4-x)/(x^(1/3)(6-x)^(2/3))
And that I know is right. The second derivative is -8/(x^(4/3)(6-x)^(5/3))
I did all the work but I am not getting the right answer... I looked it over but I don't know where I am going wrong... can someone do like the fisrt several lines? Or the whole thing... Thanks!
Is there anyone that can help me with this!!!!!!!!!!!!!! PLEASE!!!!!!
d/dx((6-x)^(1/3) x^(2/3))
| Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = (6-x)^(1/3) and v = x^(2/3):
= | x^(2/3) (d/dx((6-x)^(1/3)))+(6-x)^(1/3) (d/dx(x^(2/3)))
| Use the chain rule, d/dx((6-x)^(1/3)) = ( du^(1/3))/( du) ( du)/( dx), where u = 6-x and ( du^(1/3))/( du) = 1/(3 u^(2/3)):
= | (x^(2/3) (d/dx(6-x)))/(3 (6-x)^(2/3))+(6-x)^(1/3) (d/dx(x^(2/3)))
| The derivative of x^(2/3) is 2/(3 x^(1/3)):
= | (x^(2/3) (d/dx(6-x)))/(3 (6-x)^(2/3))+(2 (6-x)^(1/3))/(3 x^(1/3))
| Differentiate the sum term by term and factor out constants:
= | (x^(2/3) (d/dx(6)-d/dx(x)))/(3 (6-x)^(2/3))+(2 (6-x)^(1/3))/(3 x^(1/3))
| The derivative of 6 is zero:
= | (2 (6-x)^(1/3))/(3 x^(1/3))-(x^(2/3) (d/dx(x)))/(3 (6-x)^(2/3))
| The derivative of x is 1:
= | (2 (6-x)^(1/3))/(3 x^(1/3))-x^(2/3)/(3 (6-x)^(2/3))
d^2/dx^2(x^(2/3) (6-x)^(1/3)) = -(2 (6-x)^(1/3))/(9 x^(4/3))-(2 x^(2/3))/(9 (6-x)^(5/3))-4/(9 (6-x)^(2/3) x^(1/3))
The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where âvâ is the initial velocity (speed) in ft/sec and âkâ is the initial height in feet (as if you were on top of a tower or building).
Sure! Let's go through the process of finding the first and second derivatives of the function f(x) = x^(2/3)(6-x)^(1/3) step by step.
First, let's find the first derivative. To do this, we can use the product rule. The product rule states that if we have a function h(x) = f(x)g(x), where f(x) and g(x) are both differentiable functions, then the derivative of h(x) is given by:
h'(x) = f'(x)g(x) + f(x)g'(x)
Applying this to our function f(x) = x^(2/3)(6-x)^(1/3), we have f(x) = x^(2/3) and g(x) = (6-x)^(1/3).
Let's find the derivatives of f(x) and g(x) separately before applying the product rule.
The derivative of f(x) = x^(2/3) can be found using the power rule for differentiation. The power rule states that if we have a function h(x) = x^n, then its derivative h'(x) is given by:
h'(x) = nx^(n-1)
Applying this to f(x) = x^(2/3), we have:
f'(x) = (2/3)x^((2/3)-1)
= (2/3)x^(-1/3)
= (2/3) / (x^(1/3))
Now, let's find the derivative of g(x) = (6-x)^(1/3). We can also use the power rule here:
g'(x) = (1/3)(6-x)^((1/3)-1)
= (1/3)(6-x)^(-2/3)
= (1/3) / ((6-x)^(2/3))
Now that we have f'(x) and g'(x), we can apply the product rule to find the derivative of f(x) = x^(2/3)(6-x)^(1/3):
f'(x) = f'(x)g(x) + f(x)g'(x)
= [(2/3) / (x^(1/3))] * (6-x)^(1/3) + x^(2/3) * [(1/3) / ((6-x)^(2/3))]
Simplifying this expression further may be necessary, but let's double-check this first derivative expression to ensure it matches the one you provided.